THE GEOMETRICAL SQUARE: WITH THE USE THEREOF IN PLAIN and SPHERICAL trigonometry. Chiefly intended for the more easy finding of the HOUR and AZIMUTH. BY SAMUEL FOSTER, Sometimes Professor of Astronomy, in GRESHAM colleague, LONDON. LONDON, Printed by R. & W. LEYBOURN. M. DC. LIX. A DESCRIPTION OF THE SQUARE. THE whole superficies is divided into four lesser Squares, by the Diameters FG and HI. Each of the 4 Semidiameters of, EH, EG, EI, are divided as the lines of Sines upon the Sector, the Semidiameters being the whole Sine, And through the parts of each Semidiameter are drawn right lines perpendicular thereunto, quite over the face of the whole Square every 10th, 5th, &c. are to be distinguished from the rest, for the more easy and speedy account. Upon the limb are inserted several Scales, for several uses. The edges of these Scales bordering close upon the sides of the inner Square, that it may be discerned which lines and parts of the Scales do butt one upon the other. On the sides AB, AC, are inscribed Scales of equal parts, the whole being divided into 10, and subdivided as quantity will give leave. The parts are numbered by 1. 2. 3. 4. 5. 6. 7. 8. 9 10, and may stand, either for 1. 2. 3, &c. or for 10. 20. 30, &c. or for 100 200. 300, &c. as occasion requireth. To the lines by and CG, are annexed Scales of right sines whose beginning is at B and C, and the end at I and G, numbered by 10. 20, &c. to 90. The other parts ID, GD, have Scales of Tangents from 0 gr. to 45 gr. numbered either from G to D, and so to I. Or from I to D, and so to G, or rather both ways, by 10, 20, 30, &c. till 90, and these divisions have respect to the centre E. Lastly, a Scale of larger Tangents, lying behind these last named, Parallel to the sides BD, CD, beginning 0 gr. from B and C, and so proceeding to 45 gr. in D and ending in 90 gr. at C and B, accounted both ways. These have respect to their centre A. {non-Roman} {non-Roman} {non-Roman} {non-Roman} {non-Roman} There is further added a thread and plummet, which is to be used in every practice, and must be in length equal to the lines of, and FG. And if the thread be found inconvenient in practice, because it will take up the use of both hands, there may instead of it, be used a little bow, the thread of it being at the least equal to AD, which will perform the office of the other thread by the help of one hand only, or a straight ruler may serve, if it be thought convenient for that purpose. If the Square be applied to the observation of Angles, it may be fitted thereto one of these two ways, Either by placing two sights upon the side of the Square, one upon the centre A, the other upon the line AB, which issueth out of the centre A. And a running sight contrived upon the utter edge of the Instrument to move from B to C by D forward, and so from C to B by D, backward again; Or else if this be thought inconvenient, or not feasible because of the sights turning over at the Angle D, then this movable sight may go only upon one of the sides BD or CD. And for that purpose the sight at A, is to stand precisely upon the centre, and both the sides AB, AG must have sights there fixed, as precisely, upon their lines that come from A. Of the use of the Square in General for the Solution of Spherical Triangles. In any Spherical Triangle whatsoever. ¶ By having the Legs and Base, to find the Vertical Angle. THe Angle given or sought is the Vertical Angle, The sides comprehending it are the legs. The side subtending it is the Base. From the top of the Square, count the sum of the legs upon one side, the difference of them on the other side, To this sum and difference apply the thread, Then from the same top of the Square count the base also, And mark where it cuts the thread, for the line passing through the intersection, and standing Square to the top, (if it be numbered from that side of the Square whereon the difference of the legs was counted) gives the Vertical Angle required. This is the general manner of work for this Proposition, which may be illustrated by these particulars. FIRST, Having the Latitude of the place, the Declination and Altitude of the Sun, to find the Hour of the day. BY the declination of the Sun, may be had his distance from the elevated Pole, By subtracting it from 90 gr. when the Declination is of the same denomination with the said Pole; Or by adding the Declination to 90 gr. when the Declination and elevated Pole are of several denominations. In this case, we have the three sides of a Spherical Triangle given, and an Angle sought. The two legs are The compliment of the Latitude, and The sun's distance from the Pole. The base is, The compliment of the sun's Altitude: The Angle is the Hour required, which must be accounted from the Coast of a contrary name, to the elevated Pole. According then to the former general prescript, and this particular declaration, For the hour take the sum and difference of the compliment of the Latitude, and of the sun's distance from the Pole, and from the top of the Square, upon one side, count the difference, the sum on the other, to these terms apply the thread; Then from the top of the Square also, count the compliment of the sun's Altitude, and where it cuts the thread, the line that crosseth it Square in the same point (being reckoned from that side whereon the difference of the legs was counted) gives the hour from the Meridian or noon. To make it plain by an Example. In a North Latitude of 52 gr. 30 min. the Sun declining 20 gr. to the North, the Altitude of the Sun being by observation 43 gr. I would know the Hour of the day. The legs of this Triangle are the compliments of the latitude and declination, that is 37 gr. 30 min. and 70 gr. 0 min. The sum of them is 107 gr. 30 min. their difference is 32 gr. 30 min. Then from the top of the Square at D upon the side DG, I reckon this difference 32 gr. 30 min. downward to k. And on the other side of the Square from the top at B, I also count the sum of the legs 107 gr. 30 min. downward to l. To k and l. I apply the thread. Which done from the top of the Square, again, I count the base 47 gr. the compliment of 43 gr. the altitude observed, downward also to 0, and the line that there meets me, I follow till it cut the thread, which is at n, and the line that there ariseth Square to it is n r. I say now that nr, if it be counted from the side DC whereon the difference of the legs was counted, shall give 44gr. 8 min. which turned into hours and minutes of an hour, (allowing 15 gr. to an hour; and 15 min. of a degree to one minute of an hour) will make two hours ●nd 56½ min. from the Meridian or South, And such is the Hour for that Latitude, Altitude, and Declination. So also, If in the same Latitude and distance of the Sun from the Pole, but in the altitude of 10 gr. I would know The bower of the day. Here because the legs, that is, the compliment of the latitude and the distance from the Pole, are the same, therefore the same position of the thread remains still, I therefore only reckon the base (as before) which here is 80 gr. from D to p, than I follow the line p, till it cuts the thread at m, and the line there arising is m s, which counted from DC, whereon the difference of the legs was reckoned, shall give 99 gr. 50 min. that is 6 hours' 39⅔ min. of an hour from the Meridian or South. Another Example. In the same latitude of 52 gr. 30 min. let the declination of the Sun be 20 gr. to the South, where his distance from the elevated Pole is 110 gr. and let the altitude of the Sun be by observation 10 gr. I require the Hour. The legs are 37 gr. 30 min. the compliment of the latitude, And 110 gr. the sun's distance from the Pole. The sum of them is 147 gr. 30 min. The difference 72 gr. 30 min. which I count upon the sides of the Square down to u and t; and the base which is 80 gr. the compliment of 10 gr. I count also from D to p, than I follow the line p, till it cut the thread at x, and the line there arising is x y, which counted from DC, whereon the difference of the legs was reckoned, shall give 38 grad. 56 min. that is, two hours and almost 36 min. of an hour from the Meridian or South. Note, That the thread in this situation, shows on the diameter of the Square (which in this case represents the Horizon) the Semidiurnal and Seminocturnai Arks, for where the thread crosseth the middle line, the line there arising, (counted from that side of the Square, whereon the difference was numbered) shows the Semidiurnal ark, and counted from the other side, shows the Seminocturnal ark. Observe also, If you would known the Crepusculum or Twilight, the thread is to be placed as before, according to the sum and difference of the legs, and if you allow 18 gr. for the Crepusculin line (as they usually do) the base will always be 108 gr. which in the two first Examples will not touch the thread at all, and therefore in that latitude and parallel of the Sun, the twilight continues all night. But in the last Example you shall find the Crepusculin line to cut the thread, 6 hours and 15 min. from the Meridian, which shows that the twilight begins at 5¾ a clock in the morning, and ends at 6¼ in the evening, and the rest of the time is dark night which is 11½ hours. If the sum of the the legs be more than 180 gr. that is, if it would reach beyond the bottom of the Square, you must when you have reckoned to the bottom, count upward back again till you have ended the whole sum. SECONDLY, Having the Latitude of the place, the Declination and Altitude of the Sun, To find the Azimuth of the Sun. HEre also the 3 sides are given, the same with the former, and an Angle sought. The two legs are the compliments of the latitude, and sun's altitude, The base is the sun's distance from the Pole which is elevated above the Horizon. The angle sought is the sun's Azimuth, from that part of the Meridian, which is of the same denomination with the elevated Pole. So then according to the former general prescript, and this particular declaration, for the Azimuth, do thus. Take the sum and difference of the compliments of the latitude and sun's altitude, and count from the top of the Square, the one upon one side, the other on the other side, and to these terms apply the thread; Then from the top of the Square also, count the sun's distance from the Pole, and where it doth cross the thread, the line that there ariseth Square to the former, being reckoned from that side of the Square whereon the difference of the legs was counted, gives the Azimuth from that part of the Meridian which is of the same denomination with the elevated Pole, and counted from the other side, gives the Azimuth from the other coast. To Illustrate it by an Example. Another Example, In the same latitude and the same altitude, and therefore also the same situation, of the thread, let the declination be Northerly 23½ gr. therefore the distance from the Pole will be 66½ gr. which I count from D to f, and following the line f till it meet with the thread at i., I find the line g i, to cross there also, which being counted from the side DC, whereon the difference of the legs was counted, shows 79 gr. 38 min. the Azimuth from the North, Or counted from the other side, gives the residue of the former, 100 gr. 22 min. The Azimuth from the South. A third Example. In the same latitude and altitude, and therefore also in the same situation of the thread, let the declination of the Sun be 10 gr. to the South, then shall his distance from the elevated North Pole be 100 gr. and because this 100 gr. is the base, I therefore count it from the top D, down to o, and following the line o, I find it to cut the thread at r, and the line r p there crossing, shows me from DC, (the side whereon the difference of the legs was counted) 146 gr. 32 min. for the Azimuth from the North, or if the same line be numbered from the side BA, it shows 33 gr. 28 min. the residue of the former, for the Azimuth from the South. These Examples may suffice for this kind, and according to these patterns, all others are to be framed. In any Spherical Triangle whatsoever. ¶ By having the legs and Vertical Angle, to find the Base. From the top of the Square, count the sum of the legs upon one side, the difference of them on the other side. To this sum and difference apply the thread: Then from that side of the Square whereon the difference of the legs was numbered, count the Vertical Angle given, and where it cuts the thread, mark the line that passeth therethrough parallel to the top of the Square, for that line, counted from the top, gives the base required. This is general for all works of this kind, which may be illustrated in particular: thus, Having the latitude of the place, the declination of the Sun, and the hour of the day, to find the altitude of the Sun, for that latitude, declination, and hour. HEre have we the two legs of the Triangle, with the intercepted Vertical Angle, given, and the base sought. The legs are the compliment of the latitude, and the sun's distance, from the Pole; The angle intercepted is the hour whose altitude we seek. And the base is the compliment of the altitude sought for. Wherefore by the former general prescipt, and this particular explication, we may attain to the thing required thus, as in the former practice, so here; Apply the thread to the sum and difference of the compliment of the latitude, and of the sun's distance from the Pole, Then reckon the hour given from the Meridian, from the side whereon the difference of the legs was counted, and where it crosseth the thread observe the line that passeth therethrough parallel to the top of the Square, for that line reckoned from the top, shows the base, that is, The compliment of the altitude, or reckoned from the middle line of the Square, it gives the altitude itself of the Sun, for that parallel and Hour; And so the thread (which now represents the sun's parallel:) lying still, you may count the altitudes for all the rest of the hours for that parallel. For Example. In a North latitude of 52 gr. 30 min. let the Sun decline 20 gr. to the North, so that his distance from the elevated North-pole will be 70 gr. which is one of the legs given, and the compliment of the latitude, 37 gr. 30 min. is the other, The sum of them is 107 gr. 30 min. The difference is 32 gr. 30 min. This difference is the compliment of the sun's Meridian altitude; and I count it from D the top of the Square, to k: (in the first figure) thereto applying one end of the thread And on the other side from B, I count the sum, 107 gr. 30 min. down to l, thereto applying the other end of the thread. The thread thus laid, resembles the sun's parallel, for that declination, Now from the side D k, whereon the difference was numbered, I count the Vertical Angle, As first 15 gr. for the first hour from the Meridian, either 11 in the morning, or one in the afternoon, and where it cuts the thread, I observe the other line there crossing also, which counted from the top, gives for the base, 34 gr. 32 min. the compliment of the altitude required. Or rather. Count it from the middle line, which in this case represents the Horizon, and then you shall have 55 gr. 28 m. the altitude itself, for that hour and parallel, So the second hour from the Meridian (10 or 2) gives for the altitude 50 gr. 4 min. The third hour (9 or 3) gives 42 gr. 31 m. The fourth hour from the Meridian (8 or 4) gives 33 gr. 53 min. The fifth (7 or 5) gives 24 gr. 48 min. The sixth (6 in the morning and evening) gives 15 gr. 45 m. The seventh (5 in the morning, or 7 in the evening) gives 7 gr. 6 m. And so farrethe Sun is above the Horizon in that parallel, and then begins to go down. And observe further, That the thread thus placed taken in that part below the Horizon, gives the altitudes for the hours in the declination which is equal to this, but to a contrary coast; so that the thread in this situation, gives the altitudes for the declination of 20 grad. towards the South, for that part of the thread that is under the Horizon or middle line, is the seminocturnal ark for the parallel lying 20 gr. from the Equinoctial Northward, and is therefore equal to the Semi-diurnal ark that belongs to the parallel which lies 20 gr. from the Equinoctial Southward, and is of like situation below the Horizon that the other is above, wherefore the depressions belonging to the hours in this, are the same with the altitudes of the same hours in the other. To go on then where we left, The next hour counted from the Meridian of the Winter parallel is the fourth, that is, either 8 in the morning, or 4 in the afternoon, and his depression is 0 gr. 50 min. The next hour the third from the Meridian (either 9 or 3) is depressed 7 gr. 39 m. The second hour, (10 or 2) is 12 gr. 57 m. The first hour (11 or 1,) is depressed in this North parallel 16 gr. 20 min. that is, it is elevated so much in the like South parallel. Thus of each two opposite parallels of declination may the altitudes be had at one and the same situation of the thread. But if the other way seem plainer, do as before. Let the Sun in the same latitude decline 20 gr. to the South, his distance from the North elevated Pole, is then 110 gr. the sum of the compliment of the latitude 37 gr. 30 min. and this distance is 147 gr. 30 min. The difference is 72 gr. 30 m. This difference I count from D to t, (in the first figure,) The sum I also account as before, from B to u, And to t u, I placed the thread, Now from the side D t, I count the hours as I did before, and find the altitude of 11 and 1, 16 gr. 20 min. of 10 and 2, 12 gr. 57 min. of 9 and 3, 7 gr. 39 min. of 8 and 4, 0 gr. 50 min. All the same that the former depressions were; And if now you take the depressions of the hours upon the thread in this situation, you shall find them all the same that the altitudes in the former parallel of 20 gr. North declination were; So that ever, one side of the thread will afford the altitudes for the hours in any two opposite parallels. The Meridian altitude is the compliment of the difference of the legs, And in the opposite parallel it is the excess of the sum of the legs above 90 gr. And as you have done for the Altitudes of the whole hours, so may you do for their halves and quarters. Thus much for this also. In any Spherical Triangle, whatsoever. ¶ If the Proportions be in right Sines alone, they are resolved in this manner. Count the first sine given (upon one of the sides of the lesser Square EIDG,) from the centre E, and upon the line there arising count the second sine, whereto apply the thread, Then upon the same side with the first, count the third, and observe the line there arising, for from it doth the thread cut off the fourth sine required. This general may be illustrated in particular thus. Having the greatest Declination of the Sun, and his distance from the next Equinoctial point▪ to find the Declination of the Sun for that distance. THis particular belongs to the solution of a Rect-angled Spherical Triangle, yet the manner of the work in this is the same with the work belonging to the solution of the Obliquangled ones. The proportion stands thus; As the radius, is to the sine of the greatest declination; So the sine of the sun's distance from the next Equinoctial point, to the sine of the declination of that point. For an Example. Let the distance from the Equinoctial be 30 gr. The greatest declination 23 gr. 30 min. I would know the declination for that distance of 30 gr. The Proportion is, As the radius, is to the sine of 23 gr. 30 min. So the sine of 30 degrees, to what fine? Count EG for the radius, and upon GD the line there arising, reckon 23 deg. and 30 min. up to a, and thereto apply the thread, Then again upon EG, count e b, the sine of 30 gr. and follow the line there arising which is b c, till it cut the thread at c, and the line c d, there crossing also (being counted from EG) gives for the declination required, 11 gr. 30 min. So that the sine of 11 gr. 30 min. is the fourth Proportional Sine to the former three. By the Hour of the Day given, with the sun's distance from the elevated Pole, and the compliment of his altitude above the Horizon, to find his Azimuth. ¶ And here also is to be noted, That when any Proportion in right sines alone is offered, and the radius is the first leader in the Proportion, that then I say, it may be resolved by the former kind of work, by the sum and difference, counting the compliment of the arks of the two Sines given, for the legs of the Triangle, and the ark of the radius or 90 gr. for the Vertical Angle, and the base found out to be the compliment of the ark required. As in the first Example, The two middle arks were 30 gr. and 23 gr. 30 min. their compliments are 60 gr. and 66 g. 30 min. The sum of these is 126 gr. 30 min. there difference 6 gr. 30 min. to this sum and difference, I apply the thread, as in the former Examples, and then count the Vertical Angle 90 gr. which falls in the middle line and where the thread cuts it, there is the quantity of the Declination, 11 gr. 30 min. as before: And these degrees are counted from the centre of the Square at E. And thus may all others of this nature, having the radius in the first place, be absolved. And not only these of sines alone, but with sines intermingled with Tangents also, If it so fall out that these Tangents be less than the radius, And if instead of their proper arks be taken the compliments of the arks of sines equal unto those Tangents. And thus much for Exemplifying in this kind also. Those that follow are appropriate to rectangled Spherical Triangles only. In any Rectangled Spherical Triangle whatsoever. ¶ If the Proportion stand between right sines (whereof the Radius is always one) and Tangents, they are to be resolved in this manner. Upon one of the sides of the lesser square EIGD. Count the first term, and upon the line there arising count the second, whereto apply the thread. Then upon the same side whereon the first was reckoned, count the third, and follow it till it cross the thread, for the quantity of it comprehended between the third term and the thread, gives the fourth proportional term required, always remembering that every term be taken on his proper Scale. Here because the proportions are divers, we shall need more explication then in all the rest. Yet the variety herein, may be reduced to three ways according as one of these three, either the Radius; Sine, or Tangent, doth lead in the Proportion, the three ways are these: 1 As the radius, is to a tangent, So is a sine to a Tangent. 2 As a sine, is to a tangent, So the Radius is to another tangent. 3 As a tangent, is to the radius, So another tangent, is to a sine. But this variety is not all, for each of these three ways is subject to variation, and that upon this occasion.— upon the square we have no tangent greater than the radius, or tangent of 45 degrees. Wherefore the proportion must be so contrived, as that no tangent greater then of 45 gr. be ingredient into it. To that purpose serves this general direction, namely,— If the tangent which is copartner, in the proportion with the sine, be greater than of 45 gr. (Always provided that the two tangents do never stand immediately together, which if they do, may be brought into frame by transposition or alteration of the middle term.) Then, In the two first ways the radius and sine must change places; and for the two tangents must be taken the tangents of their compliments; In the third way, the cotangents of the third and first terms must remove into the first and third places. To show this more particularly in the 3 former ways. In the first, If the tangent required in the fourth place prove greater than of 45 gr. (which how to discover is showed hereafter) then by the former direction this alteration must be made. As the sine in the third. place, is to the cotangent in the sesecond, So is the radius in the first place, to the cotangent of the fourth. In the Second, If the tangent in the second place, be greater than of 45 gr. then by the former direction this proportion must be thus changed. As the radius in the third place, is to the cotangent of the second. So is the sine in the first place, to the cotangent of the fourth. In the third If the tangent in the third place, be greater than of 45 gr. then according to the former prescript this proportion must thus be varied. As the cotangent of the third place, is to the radius in the second; So the cotangent of the first place, is to the sine required in the fourth place. Because in the first proportion it is unknown whether the tangent required in the fourth place be greater or lesser then of 45 gr. and yet is necessary it should be known before it can be found out, you shall therefore in practice discover it thus. If the line whereon it is to be accounted doth not meet with the thread rightly situated upon the Square, then is it greater than of 45 gr. and then the proportion must be altered as before, but if it do meet with the thread, than is it less than of 45 gr. as it should be. Observ● that the tangents are actually in the limb only, yet may be understood to be all over the plain, for some line or other standing even against them in the plain will supply them as well as if they were actually there drawn. And note that if such a Proportion as this do at any time happen, namely, As a sine is to a tangent, So another sine, is to another tangent. And that these tangents, be discovered to be one of them greater than of 45 gr. the other less, That then the radius is to be brought into the Proportion, by saying, As the first sine, is to the first, So is the radius to another Tangent; Then leaving out the first sine and tangent, and using for them the radius and this later tangent, say, As the radius is to the last found tangent, So is the sine in the third place to the tangent in the fourth; Which Proportion suits with those going before. But if both the tangents be either greater or less●r then of 45 gr. then may the solution be made without the help of the radius. According to the former Rules generally delivered are these following Examples framed, and will fully illustrate every Case. For the first of the three several ways there are three cases, For either both the tangents are less than 45 gr. or both greater, or no less, the other greater. 1 As the radius is to the tangent of 40 gr. So the sine of 50 gr. to the tangent of what? upon the Square EIDG, I count EG for the radius, and upon the end of it in the Scales of tangents I reckon G a, 40 gr. thereto applying thread. Then upon EG, I count the sine of 50 gr. from E to b, and follow the line there arising till it cut the thread E a, at c, So that b c is the fourth term required, which because it is a tangent, must be numbered in the Scale of tangents, and therefore by help of the line c e, I transfer it thither, and find that the line lies even against 32 gr. 44 min. in the Scale of Tangents, and this is the ark of the fourth proportional term required. 2 As the radius is to the Tangent of 60 gr. So the sine of 50 gr. to the Tangents of what? As the sine of 50, is to the cotangent of 60, So is the Radius, to a fourth Tangent, which will be the compliment of that which should be produced by the former Proportion. By this alteration it comes to pass that the sine leads in the Proportion, and so this Example now falls under the Examples of the second General way, and therefore shall be resolved there. 3 As the radius is to the Tangent of 50 gr. so the sine of 50 gr. to the Tangent of what? Upon the Square I take EG for the radius, and at the end of it I reckon up to D, which is 45 gr. and so forward on the other side to g, that is to 50 gr. Then upon EG, I count e b the fine of 50 gr. and follow the line there arising till it cut the thread at i., so that b i. is the fourth term, and because it is a Tangent, therefore by help of the line passing through i., that is, by the line i m, I transfer it to the Scale of Tangents, and find that lies even against 42 gr. 24 min. which is the fourth ark required. For the second of the three general ways, there are two Cases; For the Tangent that is Copartner with the sine in the Proportion, may be either lesser or greater then of 45 gr. for the lesser, take the Example which before was preferred hither, namely, 1 As the sine of 50 gr. is to the Tangent of 30 gr. So the radius is to the Tangent of what? First, upon the side EG, I count the sine of 50 gr. and to the line there arising, I transfer Gt the Tangent of 30 gr. by help of the line t s, and to s, I apply the thread, which thread cuts the limb in u, so that G u I find to be the Tangent of 37 gr. and this is the fourth term required in this Proportion; But in the second Example going before, whereof this is also the solution, this 37 gr. is the compliment of the fourth ark there required, so that the fourth ark there, should be 53 gr. which because it is greater than 45 gr. is therefore absolved this way, and not the other. 2 As the sine of 50 gr. is to the tangent of 50 gr. So the radius is to the Tangent of what? Here because the Tangent of 50 (being copartner in the Proportion with the sine) is greater than the Tangent of 45 gr. and so cannot be expressed upon the square, therefore the Proportion must be altered by changing the places of the first and third terms, and by taking the compliment of the second and fourth, after this manner. As the radius is to the Tangent of 40 gr. So the sine of 50 gr. to the Tangent of 32 gr. 44 min. the compliment whereof 57 gr. 16 min. answers to the question in the former Proportion, and this last Proportion falls under the first general way where the radius leads, and was resolved before in the first practice upon the Square, As EG, to G a, So E b, to b c, or G e the Tangent of 32 gr. 44 min. In the third of the three general ways, there are two cases, according as the Tangent of the third place, which is copartner in the proportion with the sine, is lesser or greater than the Tangent of 45 degrees. 1 As t●e tangent of 40 gr. is to the Radius, so the Tangent of 32 gr. 44 min. to what fine? Upon the side GD, I count G a, the Tangent of 40 gr. and thereto apply the thread, then upon the same side GD, I reckon also the Tangent of 32 gr. 44 min. from G to e, and follow the line meeting at e, till it cut the thread at c, and the line there crossing also is c b, which counted from e, the centre of the Square, gives the sine of 50 gr. which is the fourth term required. 2 As the Tangent of 60 gr. is to the radius, So the Tangent of 53 gr. to what fine? Here because the Tangent of 53 gr. being copartner in Proportion with the sine, is greater than of 45 gr. therefore the first and third terms must change places, and their compliments are also to be taken, thus, As the cotangent of 53 gr. or Tangent of 37 gr. is to the Radius, So is the cotangent of 60 gr. or Tangent of 30 gr. to the fourth sine required. Upon GD the side of the Square, I count G u the Tangent of 37 gr. thereto applying the thread. Then on the same side of the Square, I also count G t, the Tanegnt of 30 gr. and follow the line at t, till it cut the thread at s, and the line s b, there crossing being counted from e, the centre of the Square, gives me the Sine of 50 gr. the fourth term required. These Examples are sufficient to give light to the rest, For no Proportion can fall out in these kinds, whereunto these Proportions and their Examples are not suitable. And so much of Spherical Triangles. Of the use of the Square, in Right-lined Triangles. IF the Proportions be between Tangents and equal parts, then are we to use the equal parts on the sides AB, AG, as also the larger Tangents upon the two other sides of the Square, and then the work will be the same, for form, that was before in Tangents and sines, for the lines on the superficies will carry the parts of either of these Scales to and fro, as they did before the parts of the Scales of the lesser Tangents. If the Proportions be between sines and equal parts, then are we to make use of the sines inscribed upon the Scales by, CG, together with the former equal parts, the lines upon the superficies still acting their former parts of carrying from the one to the other. Examples in these kinds, And first of sines, and equal parts, or Numbers. SUppose at the two stations DC, I had observed the angles BCA, 30 gr. BDA, 50 gr. and CD the difference of Stations 40 feet, and by these observations, I require to know the altitude AB. First, I must find the length of the lines CB, or DB, in this Example of CB, after this manner, because BCA is 30 gr. and BD A 50 grad. therefore their difference CBD is 20 gr. Now then, As the sine of CB D 20 gr. To the length of CB required. To resolve this upon the Square, from C, I count the line of 20 gr. to a, and observe the line there meeting me, then upon the side AC, I count A d 40 equal parts or feet, and thirdly, I reckon C c the third term, which is the sine of 50 gr. and follow the line there meeting me, till it cross the thread (which was to be applied to b, the intersection of the lines a b, coming from the first term, and d b rising from the second) at e, and there I find another line concurring, namely, e h, which I follow down to h, and there it shows in the equal parts A h 89 feet, and 58 centesmes or hundredth parts of a foot, And this is the length of CB, now to get BA, by a second work, I say, As CB the radius, to BA the sine of BCA 30 gr. So BC 89 58/100 feet, to BA the altitude in feet. To perform this Proportion, Upon the Square I take AH, equal to by the radius, and upon HE, I count H oh, equal to C r the fine of 30 gr. thereto applying the thread; Then from A to h, I count the length of CB, that is 89, 58, and so follow the line there arising, up to the thread to s, where I find the line s u, limiting out A u, 44, 79, that is 44 feet, and 79 telesmes of a foot, and such is the altitude of AB required. Thus by having the three Angles of a plain Triangle, and one side you may find the two other sides; And by having two sides and an Angle opposite to one of them, you may find the other two Angles and third side, in any Right-lined Triangles whatsoever. Examples of equal parts, and Tangents. This kind of work may sufficiently be explained in the solution of this problem. ¶ By having an Angle and the two sides comprehending it, to find the other Angles. First, if the Angle comprehended be a Right-angle the work is easy. As AB 30, to BC 20 So AB the radius to BC, The Tangent of BAC. Therefore upon the Square I count A w 30 equal parts, and follow w f, till it stand even with 20 equal parts counted on the side AB, and laying the thread at f, I find it to cut in the limb of the greater Tangent C y, which is 33 gr. 41 min. And such is the quantity of the angle CAB. And the compliment of it 56 gr. 19 min. is the quantity of the angle ACB. Further more it is to be noted, That if by having the right angle with the two including sides, you would find the sub tending side AC. In this case one of the acute angles must first be sought, and then by the Proportions of sines and equal parts, the side AC may be had. So also, If by having the distance AB 30 foot, and the angle CAB 33 gr. 41 min. I would know the Height BC, Upon the Square I lay the thread from C to y the Tangent of 33 gr. 41 min. then upon the equal parts I count A w 30, & follow the line rising at w, till it meet with the thread at f, and at f, I find the line f q crossing also, which followed to q, shows in the limb 20 equal parts for the altitude BC. By these mixed Proportions of equal parts with sines and Tangents, may all mensurations be performed, as also all conclusions upon the Common Sea-chart, with Mr. gunter's corrections of it, to make it sufficient for sea-mens' use. Secondly, in any plain Triangle whatsoever. The former problem may be resolved in general by this Proportion, As the sum of the two sides, is to their difference, So is the Tangent of half the sum of their opposite Angles, to the Tangent of the half difference of those Angles. As 60 the sum of the sides, is to 20 their difference, So is the Tangent of 30 gr. the half sum of the angles at A and C, to the Tangent of their half difference. The best way for the solution of Proportions in this kind is first (as was before admonished in the joint use of Sines and Tangents) to seek out a Tangent whereon the Radius is in Proportion as the sum of the legs is to the difference of them, which Tangent is ever less than the Tangent of 45 gr. or radius, because the difference of the legs is always less than the sum of them. And when the radius is brought in, the Proportion may be absolved upon the lesser Square, which is fitted for the Proportions of Sines and Tangents, in the same manner as was showed in the like Examples before. And the Proportion will then stand thus. As the radius, is to this new found Tangent, So is the Tangent of half the sum of the angles, to the Tangent of half their difference. To make it plain by the former instance, As 60 parts, are to 20; So is the radius, to what Tangent? Upon the Scale of equal parts, I account A k 60 and on the line there arising I account k x 20, thereto applying the thread, and then I see it cut off in the greater Tangents C 18 gr, 26 min. which is the Tangent sought. And now that the radius is brought in, the next Proportion will be thus, As the radius, is to the Tangent of 18 gr. 26 min. So is the Tangent of 30 gr. half the sum of the angles, to what Tangent? Upon the lesser Square, I take EG for the radius, and count G ζ the tangent of 18 gr. 26 min. thereto applying the thread, then upon ID, I count I x the Tangent of 30 gr. and follow the line thence passing to the thread at σ, where the line σ φ shows, in the limb the Tangent G φ, which is the Tangent of 10 gr. 54 min. the half difference of the angles required, which added to 30 gr. the half sum, makes the greater angle 40 gr. 54 min. And taken from the same 30 gr. leaveth 19 gr. 6 min. for the lesser angle. ¶ By having the three sides of a plain Triangle, to find the Angles. THe first work here will be to let fall a perpendicular, and to know where it will fall, and so reducing the Triangle to two Rectangles, you may resolve them as Rectangles, either by sines and equal parts, or Tangents and equal parts. The manner of dividing a Triangle, into two Rectangles, as also to find the place where the perpendicular falls, is showed by Mr. Gunter in the first Book of his Cross-staff, and the Proportion for the solution of it, is a proportion of equal parts or numbers only, the manner of which is hereafter showed in the next use of the Square in numbers or equal parts alone. Thus far of the use in Right-lined Triangles. Of the use of the Square in Proportions of equal parts or Numbers only. THe equal parts with the lines on the superficies to carry them along, will perform them very sufficiently and expeditely, If any number be too great take 1/10 or 1/10 part of it, and count the rest as a fraction either decimal or centesimal. And as in the former works, so here, the first & third terms, must be counted upon one side of the Instrument, the second and fourth upon the lines arising out of the terms of the former, so the thread applied to the second, will limit out the fourth. The manner of the work is always a like, and may sufficiently be declared in this one Example, I would know that number whereto 60 bears the same Proportion, that 40 doth to 50: The Proportion stand thus As 40, is to 50: So 60, to what? Upon AC the Scale of equal parts, I count AD 40, and upon the line arising out of d, (by help of the Scale of equal parts upon the other side AB) I count 50 up to l, and thereto apply the thread, Then upon AC, I count the third term, 60 to K, and follow the line there arising, till it meet with the thread at p, and there the line p m, meeting also, shows in the Scale AB at m, 75 parts, which is the fourth proportional number required. And thus in all others. Of the use of the Square in the observation of angles. WHen the observation is made and the sight placed, than the thread from A, applied to the running sight, will express the angle in the larger Tangent, And for observing any altitude or depth, the thread alone, without the help of the running sight, will express the Angle, if the observation be made as usually it is by other Instruments.— The Square at the greatest cannot observe an Angle that is greater than 90. If therefore such an Angle come to be observed, you must observe the residue of it, which is his compliment to 180 degrees. Hitherto we have had a general view of the use of the Square in all Triangles and ordinary Proportions in Numbers. Now remains the bringing of it down to particulars in every kind; which would be an infinite labour, and unnecessary to those that are any thing experienced, in the use of Instruments, especially seeing we have here a taste of every of them, and the particular Proportions are everywhere extant. Hereafter I may add something more on the other side, for the present I here make stay, and content myself with that which hath already been delivered. FINIS.